DYN3701 Assessment 1 Solutions Due 2026 |Dynamics|

UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology


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DYN3701: Dynamics of Machines

Assessment 1 — Semester 1, 2026

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DYN3701
Module Code:
Dynamics of Machines
Module Name:
Assessment 1 — 2026
Assessment:
40
Total Marks:




Submitted in partial fulfilment of the requirements for DYN3701 — UNISA 2026

,UNISA | DYN3701 Assessment 1 — 2026



Question 1: Unsymmetrical Bending of a Cantilever Beam [10]


Question


A wooden cantilever beam with a rectangular cross section and a total length of L supports
an inclined point load P at its free end, as illustrated in the figure. You are required to deter-
mine the orientation of the neutral axis resulting from the inclined loading, and then calculate
the maximum tensile stress in the beam caused by the applied load P .

The beam properties are as follows:

• Width, b = 60 mm
• Height, h = 100 mm
• Length, L = 1.2 m = 1200 mm
• Applied load, P = 450 N
• Load inclination angle, α = 27◦

Determine:

1. The orientation of the neutral axis.
2. The maximum tensile stress in the beam.


Solution


Step 1: Resolve the Load into Components


The inclined load P is resolved into two perpendicular components acting along the y- and
z-axes:



Pz = P cos α = 450 cos 27◦ = 450 × 0.8910 = 400.95 N




Py = P sin α = 450 sin 27◦ = 450 × 0.4540 = 204.30 N




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,UNISA | DYN3701 Assessment 1 — 2026



Step 2: Calculate the Bending Moments at the Fixed End


For a cantilever beam, the maximum bending moment occurs at the fixed support. Each load
component produces a moment about the respective axis:

Moment about the y-axis (caused by Pz ):


My = Pz · L = 400.95 × 1200 = 481,143.52 N mm



Moment about the z-axis (caused by Py ):


Mz = Py · L = 204.30 × 1200 = 245,154.87 N mm



Step 3: Calculate the Second Moments of Area


For a rectangular cross section, the second moments of area about the centroidal axes are:

About the y-axis:

hb3 100 × 603 100 × 216,000
Iy = = = = 1.80 × 106 mm4
12 12 12


About the z-axis:

bh3 60 × 1003 60 × 1,000,000
Iz = = = = 5.00 × 106 mm4
12 12 12


Step 4: Equation of the Neutral Axis


For unsymmetrical bending, the general stress equation is:

Mz My
σ=− y+ z
Iz Iy


At the neutral axis, σ = 0, therefore:

Mz My
− y+ z=0
Iz Iy




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, UNISA | DYN3701 Assessment 1 — 2026


Rearranging:
z Mz /Iz
=
y My /Iy

Substituting the calculated values:

z 245,154.87/5,000,000 0.04903
= = = 0.18343
y 481,143.52/1,800,000 0.26730


Step 5: Orientation of the Neutral Axis


The neutral axis makes an angle θ with the y-axis, where:

z
tan θ = = 0.18343
y




θ = tan−1 (0.18343)




θ = 10.4◦ from the y-axis


The equivalent angle measured from the z-axis is:


90◦ − 10.39◦ = 79.6◦ from the z-axis


Key Distinction
Neutral axis orientation: In unsymmetrical bending, the neutral axis does not co-
incide with either principal axis unless the load is applied exactly along one of them.
Here, the inclined load causes the neutral axis to rotate to 10.4◦ from the y-axis (79.6◦
from the z-axis), implying the maximum stresses occur at corners that are furthest from
this inclined neutral axis.



Step 6: Calculate Stresses at the Four Corners


The four corners of the rectangular section have coordinates:


y = ±50 mm, z = ±30 mm


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